 Well, good morning. In today's class, we will try to extend what little we learned for a constant Mach number shock which we did in the last class to be able to predict the motion of a strong blast wave. Let me rephrase what we have done. In the last class, we said shock moves in a medium whose pressure is P0, density is rho 0, temperature is T0. Behind the wave, the particle moves with a velocity u, the shock velocity is rs dot and what did we do? We predicted the value of P, rho and T and also the velocity behind the shock wave. Mind you, the shock is moving at rs dot, the particle is moving with velocity u, u over here and in the frame of reference of the shock stationary, what did we get? We got the medium moving towards it with a velocity rs dot or equivalently with the Mach number ms. The upstream properties were P0, rho 0, T0 and the sound speed in the undisturbed medium was A0. The Mach number behind in the frame of reference of the shock was m, the properties were P, rho and T. This is what we did. We were able to get the properties of pressure, density, temperature and the velocity in the frame of reference of the shock that is shock stationary, how particles are moving back. We were able to get these in terms of P0, rho 0 and T0. This is what we did. Let us write one or two of those expressions down. We had rho by rho 0 that is the density behind the shock to the upstream density was equal to gamma plus 1 divided by gamma minus 1 by 2 by the Mach number of the shock. That means, we are looking at Mach number ms is equal to rs dot divided by A0 over here. Therefore, we were able to get the density. Similarly, we got the ratio of P divided by rho 0 into rs dot square was equal to 2 over gamma plus 1 and then let me put the number over here minus gamma minus 1 divided by gamma into gamma plus 1 into 1 over ms square. If the value of rho by rho 0 is known, we were also able to get the velocity. That means, for the velocity m into the value of A which is equal to u1, we were able to get the value of u1 divided by rs dot which is just the inverse of this from the mass balance. We got it as equal to gamma minus 1 plus 2 over ms square divided by gamma plus 1. These are the relations which we obtained. We want to use these relations to be able to predict some motion of a blast wave and how do we do that? Well, a blast wave is something which travels with varying velocities. That means, the Mach number is continually changing in a blast wave and that creates some problems. But to be able to do that, let us examine these three expressions in somewhat greater detail. When we looked at this particular expression namely the density ratio, what did we get? We got the value of rho by rho 0 as a function of ms square. We said when the wave has a Mach number 1, the density ratio was 1. When the Mach number was very high, let us say infinity, the density ratio from this particular expression was gamma plus 1 divided by gamma minus 1 because this is very high. It comes out to be gamma plus 1 divided by gamma minus 1 at a very large Mach number. Let us put down the values at 2 or 3 intermediate values. Let us put down the value at Mach number 5, may be at Mach number 6, that means 4 square, 5 square and 6 square. That is Mach number of 4, 5 and 6. When I put it down, what do I get? I get the value of gamma plus 1, gamma minus 1 by 2 by m square at 4 square. It is 2 by 16. It is 0.185. Gamma of 1.4. gamma is 1.4, 0.4 plus 0.125, 0.625, 0.4 divided by the value of 0.4 plus 2 by 16 over here. It gives the value over here and this value will work out to be 4.57. The value for Mach number 5, let me get a color chalk over here, comes out to be around 5. The value at 6 works out to be 5.27 and this value as we saw yesterday was 6, that is 2.4 divided by 0.4, which is 6. Therefore, if I were to plot the variation, what is it I get? The variation is something like this. That means, beyond a value around 4.5 or so, the value does not really change very significantly. The value of rho by rho naught is nearly a constant value after this particular value. Let us take the value of let us say P by rho naught rs dot square similarly. Let us try to plot it again. Let us say I have the pressure behind the shaft divided by the free stream density divided by the shock velocity square over here. Now, if I look at the expression for the condition when ms that is the Mach number is very large, that means ms is infinity, this term drops out. The value is 2 over gamma plus 1. Well, ultimately the value I get is 2 over gamma plus 1. This is the value at very large value of ms over here or let us say I am plotting with respect to ms square. Then, if the ms is slightly smaller, then what happens? The value is going to get subtracted from this value. Let us see what is going to be the value. Let me plot. When the ms is 1, therefore ms square is 1, the value is 0.7. We found that when ms is 1, it is 2 over gamma plus 1. Therefore, minus gamma minus 1 divided by gamma into gamma plus 1, this gave you the value 2 gamma. It became gamma plus 1 divided by gamma plus 1 into gamma that is equal to 1 over gamma. The value came out to be 0.7. We did this yesterday. When the value is around 3 that is the value of ms is equal to 3, the value was 0.82 for a gamma of 1.4. 2 over gamma plus 1 is equal to 0.83. When the value of ms is equal to 4, the value is 0.822. Therefore, we also find, even I do not need to work anything, it is 0.82, 822, 833. This is 0.7. Therefore, the value goes like this. In other words, even for anything exceeding around 3, 3.5, the value of p0 divided by rho 0 r square is a constant. Similarly, if I look at the value of the particle velocity behind the shock in the frame of reference of the shock divided by the shock speed in rs, I get u1 divided by rs dot. Now, I plot let us say ms over here. I just should get the inverse of this because it is going to be the opposite over here. That means this is just the inverse of this. Therefore, if 1 is over here, the curve should be like this. Let us now put down the value. The value at 4 is around 0.22 at Mach number of 4, at Mach number of 5 it is 0.2, at 6 it is 0.19 and for the condition that it is ms is infinity, if I look at this expression, this becomes 0, it is gamma minus 1 divided by gamma plus 1 which is equal to 0.183. That means beyond this it is something like it is drooping, it comes down and it is almost a constant over here. That means beyond a certain Mach number, the value tends to a constant, tends to a value around gamma minus 1 divided by gamma plus 1. The value of p by rho naught r tends towards this beyond a certain value of Mach number. The value of rho by rho 0 tends to a constant around gamma plus 1 divided by gamma minus 1. Therefore, if we are really interested only for Mach numbers greater than some limit somewhere in this region, somewhere in this region of Mach numbers, somewhere in this region of Mach numbers, we can very well say that the Mach number of the shock will not very significantly influence or affect the value of u1 by rs dot p by rho naught rs dot square and rho by rho naught. Let us use this fact to try to model a blast wave. That means we will try to predict the blast characteristics when the Mach number of the shock is a little high. Therefore, now we have to go back to a shock and see how the shock strength in a blast wave varies. Now, we again go to the streak diagram T by rs versus the distance. What did we find? Well, the shock or the blast wave gets started at a high velocity and it keeps on moving and ultimately it becomes an acoustic wave and here the Mach number towards the end is almost near 1, ms is 1. Here the Mach number is quite large. Therefore, in the portion like what we have may be in this portion wherein the blast wave the Mach number is greater than around 4 or something. Well, I can use these conditions and we will try to see what happens when the initial or in the near field of the explosion wherein a blast wave gets started for which let us say Mach number is greater than around 4. How will the blast wave get affected? When I talk of these things well the Mach number also influences the pressure density and temperature behind the shock wave and therefore it might be a little more difficult. Therefore, let us do this problem. To be able to do this problem I take this part again and again draw it again. Let us take a look at the blast wave. I just take the strong part of it. I have T, I have RS over here. Well, I say the shock gets started and it goes like this. I am interested in this strong blast region. Let us therefore say well at may be I have the explosion coming taking place over here. The shock keeps moving, the blast wave keeps moving. I am talking in terms of a high Mach number shock which is formed and it is continually decaying. This we found it to decay. Let us say a point A over here. When the shock is at point A that means this is point A, the radius is RS A. What happens at this radius may be the point A enters the blast wave over here. What happens when it enters? The pressure behind it increases very rapidly and what happens to the pressure? P by rho naught RS dot square. Since the Mach number is quite high is given by the value of 2 over gamma plus 1 that means this is the pressure behind the blast wave and then once the wave propagates out or once this particle has entered it is highly compressed and then it expands out and it follows a trajectory like this. That means the shock particle of gas once blast wave goes away or like for instance I have the blast wave coming over here, the shock particle is at a high pressure, the high pressure particle expands out and this we said is the Lagrangian the particle as it enters over here. If we assume that this expansion is isentropic may be we are talking of particle isentropic or something like homentropic type of an expansion that is particle isentropic but the motion need not be isentropic as we will see a little later. Similarly this wave continues to travel let us say it comes to point B the blast wave is here it has decayed slightly less but it is still greater than around 4 or 5 and then what is going to happen a particle which enters at B let now the distance of the shock from the origin is RSB it enters over here this particle enters over here and this particle enters when the shock is slightly lower but it is still strong and therefore it also expands once it enters because it is highly compressed and it expands thereafter. Therefore what we find is may be different particles as they enter the shock wave they expand out and if I am interested in let us say this particular particle the properties for the properties may be in this particular region that means when the shock let us say let me slightly qualify this. Supposing let us say I am interested at this particular time, time is equal to T1 the shock is here in other words the shock is over here let us for the present assume that the shock is spherical therefore may be let us say this is the origin the shock is something like this over here and now what is going to happen the shock is over here I am interested in the property over here therefore the property over here is something which corresponds to this gas which is a particle gas which enters at a as it expands this is also being processed by the lead shock over here I am interested in this particular one therefore the question is can I express the property at this particular point which let us say is at a distance r from the origin from the center when the shock is at a distance rs away. I think this requires some element of further deliberation because all what I am trying to say is well you have an explosion cost over here a spherical blast wave or a lead blast wave is something over here the lead blast wave at a given time T or T1 is at a distance rs what is happening the lead shock wave compresses the gas from rho 0 to rho and then that means it compresses the gas over here let us put it down in terms of a figure let us say rho by rho 0 over here you have the lead shock you have the ambient density is rho 0 that means I have the value 1 over here this is the ambient density ahead of the shock and when the shock reaches this point well it compresses the shock to a higher value and this value since it is since the Mach number is greater than around 4 can be approximated as gamma plus 1 divided by gamma minus 1 and thereafter maybe at any distance this is equal to rs any distance r well the gases are expanding and therefore it is to be expected that maybe the value of the density will be little lower that means we can sort of assume that I can have something like a profile like this which is the value of rho by rho 0 and this is from 0 at any particular radius or rather I can say density at any particular radius I can express as a function of density at behind the lead shock divided by r plus rs and this could be anything could be linear could be exponentially dropping could be a large value I can say let it be denoted by q and such type of profile sort of power law profile for density has been used earlier in some analysis and it leads to good results or it is very illustrative of what is happening the people who did this were poor cell he did this while working at the Los Alamos science lab at Caltech this was also followed by several other people namely poor cells work was followed by work by professor Bach at Megal University followed by John Lee again and a number of people worked on this namely with the power law assumption and they were able to sort of model the blast characteristics of a strong wave let we will do something similar but we will do something which is quite preliminary not the advanced level of work which poor cell may be he was followed by Zachar on this let us try to do this problem all what I am trying to say is let us now represent let us consider only a spherical lead shock let us assume that the lead shock originates from an explosion over here at the center and let us assume that the wave is spherical the shock distance is rs over here and when it is at rs we are telling that well the Mach number of the shock is still high therefore if this is my value if I were to plot the value over here of the density the initial density ahead of the shock is rho 0 here there is a jump in density and this jump in density because the shock strength is quite high is equal to gamma plus 1 divided by gamma minus 1 into rho 0 over here this is the value and then because of the expansion behind the shock well the density keeps falling and this particular profile that means this is equal to rs this is equal to the value we say rho by the at the surface rho s is equal to r by rs this is at any radius let us say this is at any radius r e to the power q is what we are assuming with this assumption let us try to predict something like the motion of a blast wave let us try to solve the mass conservation equation and see if we are able to deduce anything further about the blast which is propagating and our interest let us be very clear on the streak diagram when we have t versus rs and the blast wave is propagating out we wish to find out what are the characteristics of this initial region which we say is the strong blast region and this is what I will do in the next 5 or 10 minutes to be able to do that let us put down the problem a little more clearly well we tell ourselves well the density at any any any distance r when the shock front is at rs is given by the density at the shock front divided by into we said r by rs to the power q this is the equation 1 we assume that is a power law now we want to look at the mass conservation equation what is the mass conservation equation well we had the explosion over here the lead shock is at a distance let us say rs from the source over here what is the total mass which is contained by the lead shock wave well the lead shock wave starts from the explosion site of the explosion keeps moving forward it has moved a distance of rs we assume it to be spherical mind you this is only for the spherical case which I am doing it could be done for cylindrical it could be done for planar what is the total mass contained let us assume that the mass of the explosive is quite small and we will not consider it we assume that when it is at radius rs the total mass here when the ambient density is rho 0 the total mass is after all the mass which was initially available and for a spherical diameter of radius rs it is equal to the mass is equal to 4 upon 3 pi into rs cube into the value of rho 0 this is the mass which is contained now this mass is the total mass and in practice what did we tell ourselves well this is the lead shock now the in practice what is happening the energy dispersion from this explosion point overall changes the density pattern what happens is at the front you have a large value of density let me put it down over here at the value of rs the density jumps from rho to rho s and then it keeps falling let us consider may be a particular radius r small r that means this radius is rs let me consider a small value of radius r and around this radius r let us consider a small element of thickness dr in other words I am interested in what is happening in this element I will try to find out what is the mass which is contained within this element which is between radius small r to r plus dr let us write it out this is r the inner radius the outer radius is r plus dr therefore the element of mass which is contained within the small annular spherical passage between these two can be written as let us let us write the expression for that dm that is the elemental mass in the particular shell is equal to the surface area 4 pi r square into the volume which is dr in the particular shell the density is rho at a distance r and mind you this r is within rs over here this is equal to dm which is contained over here you will also recall we had the density at the distance r is equal to the value was given by the ratio of the density at the surface into we said it is equal to r by rs at the surface at the lead shock radius divided by q therefore we substitute this expression into this value here and get the value of the elemental mass in this small spherical shell of width dr between r to r plus dr for whose for which case within this elemental shell the density is rho r we get the value is equal to 4 pi into r square into I get rho s I substitute the value of rho r I get rho s into r into rs to the power q into dr over here let us simplify this therefore I get this is equal to 4 pi into the value of r to the power q plus 2 into rho s divided by rs to the power q into dr this is the value of the elemental mass dm over here now if you look at the value of the density at the surface of the lead shock we told ourselves well rho s by rho 0 is equal to gamma plus 1 divided by gamma minus 1 in the limit of strong shock because what happens is you had the term 2 over ms square which for ms large it knocks out over here therefore this becomes equal to gamma plus 1 divided by gamma minus 1 and if I were to substitute the value upstream over here I get the value of dm is equal to 4 pi into I get rho 0 into gamma plus 1 divided by gamma minus 1 that was the value of rho s over here into r to the power q plus 2 into dr and of course I also have the value of rs to the power q over here I want to solve this and therefore what is the total mass which is enclosed by the sphere that is by the lead shock of radius rs well I have to integrate it out between r is equal to 0 to r is equal to rs and if I integrate it out what is it I get let me write the value here which is equal to 4 pi into rho 0 into gamma plus 1 divided by gamma minus 1 into now I get r to the power q by q plus 2 I integrate it out it becomes r to the power q by 3 that means r to the power q plus 3 divided by q plus 3 into 1 over rs to the power q is the integral and for 0 it becomes 0 for rs well this r will become rs and therefore the value of the mass which is enclosed by the lead shock wave is equal to 4 pi rho 0 r gamma plus 1 gamma minus 1 rs to the power q plus 3 divided by q plus 3 divided by rs to the power q which I can also write as equal to the value if I were to write I get 4 pi let me write it over here 4 pi into rho 0 into gamma plus 1 divided by gamma minus 1 into here I have 1 over q plus 3 into rs to the power q plus 3 over here we had r q plus 3 over here we have rs q therefore I have rs q over here therefore I have 4 pi rho 0 gamma plus 1 divided by gamma minus 1 into rs to the power q divided by 1 by 3 and what is that total mass which is enclosed by the lead shock wave after all the lead shock wave has processed the gas which was initially at a density rho 0 to a distance of rs therefore this will be equal to the value of 4 pi into rs q divided by 3 this is the volume into rho 0 and this will be identically equal to this which is equal to 4 pi divided by I take this q by 3 over here that is I get 4 pi divided by q cube by 3 rho 0 into rs cube that means 4 pi divided by q plus 3 into rs cube into I get rho 0 into the only thing which I am left with gamma plus 1 divided by gamma minus 1. Let us simplify it further I find that 4 and 4 gets cancelled on the two sides pi and pi gets cancelled over here rs cube also gets that is the lead shock radius also gets cancelled well the rho 0 also gets cancelled over here and what is it I get I get q plus 3 divided by 3 is equal to I get the value as gamma plus 1 divided by gamma minus 1. Now I can simplify this further and therefore what is it I get q by 3 plus 1 is equal to gamma plus 1 divided by gamma minus 1 minus 1 or rather the value of q is equal to 3 into gamma plus 1 divided by gamma minus 1 minus 1 and for a particular value of gamma may be for air we are considering gamma is equal to 1.4 which is equal to 2.4 by 0.4 which is 6 minus 1 5 5 into 3 is 15 I get the value of q as equal to 15. Therefore whenever we are considering let us get back to the real problem we are considering the mass enclosed by the lead blast wave and we found out we assume some density distribution starting from the lead shock the density keeps varying and what do we find if I assume the density distribution to be given by the value of density at the shock front and the density to be distribution to be given by something like a power law r by r is to the power q. I find that this value of q is around 15 which is quite a large number. Let us just discuss what this really means and it has some ramifications. Let us see what the results indicate. If I were to consider the value of rho by rho s and I am looking at the value of r by rs if I am looking at rho by rho s is equal to r by rs to the power q if q is equal to 1 well this is the type of distribution what I get that means when r by rs is at the lead shock wave which is 1 well it is distributed like this. When I have r by rs is a number is greater than 1 well it goes like this that means what is happening is since it is greater than 1 when r by rs is near to this value the magnitude is higher when r by rs is small because of the larger value this being greater than 1 the magnitude changes. As the value of q increases let us say when q is 10 it could be like this when q is 15 all what it means is well what is going to happen is the amount of mass which is being contained in the front that means let us say when q is equal to 15 what is going to happen the mass is essentially contained at the shock front there is nothing really over here all what we mean is the shock is moving most of the mass is contained there is nothing really in depth over here because most of the mass is contained over here since q is large and therefore this type of picture what we develop using for a blast wave which has a high mark number essentially tells us that most of the mass which is contained within the blast wave is concentrated at the shock front. What does this mean really see mind you we have been telling that a wave does not really transport the mass from the center and bring it over here but what we observe is whenever we are talking in terms of a blast wave the blast wave processes the gases and when the blast wave reaches the value of r s that is r s you know it is able to take the mass and sort of distribute it at the front itself rather if I were to say well my shock front is over here the shock front as it is travelling it picks up the mass and puts it over here may be it takes the mass and distributes it over the front and if the blast wave is still strong well all the mass is sort of localized gets localized at the surface it is as if the blast wave let us again plot this on the streak diagram T versus r s well the blast wave gets started once the blast wave gets started it is something like the blast wave with a broom or something it is picking up the masses locating it at the shock front and therefore all the mass which it collects in the medium gets concentrated at the shock front it is something as if it is like a broom which sweeps the air and keeps all the air at the shock front itself or something like you have snow which is there on the road the snow plow comes and picks up the snow and all the snow is accumulating at the snow plow this becomes empty over here therefore the type of picture we get is when a blast wave gets started the blast wave picks up the mass concentrates the mass at the front and in the depth that is in the zone far from the blast wave there is hardly any mass over here therefore this is the type of picture we are able to get from the power law profile and then let us pursue on it a little more and see whether what it really implies it implies that may be the front of the shock wave or the lead blast wave provided that the lead blast wave is still strong enough contains most of the mass which it is picking up and accumulating over here or it is something like a hammer all the masses contained over here it moves with a speed r s dot which is still quite high we are talking in terms of the strong blast wave and it is as if this hammer if it comes and hits a person or some building over here well it just compresses and demolishes this and this is how a strong blast wave acts. Therefore a strong blast wave in the context of the power law assumption which we made is something like a snow plow or something like a broom which sweeps the place and this type of an assumption we get may be for the mass distribution in a blast wave. Therefore we are able to picture something about the type of damages which a blast wave or a strong blast wave can do namely it picks up the mass as it moves concentrates it at the lead shock and this lead shock when it hits some building or so can cause the damage whatever happens. Therefore may be from this mass balance we are able to get some physics and the physics is well the density distribution behind the lead shock is such that when I plot the value of rho by rho 0 as a function may be the lead shock is at r s I am interested at some station any station r well at the lead shock I have a density jump and then it keeps coming and may be the type of picture what I get is not something which is gradual like this most of the density is concentrated over here and in depth I have really nothing over here that means I get a value of q that is I get q by r is equal to r by r s to the power around 15 for air may be depending on the value of gamma I could get different values this is the value of rho by rho 0 over here or you should have been at the shock front this is rho s is equal to rho by rho s over here well this is the picture of this we find therefore it can be the lead blast wave can be assumed to act something like a hammer in which all the mass is concentrated at the front therefore we tell ourselves well a strong blast wave is something like a shock wave in which the mass which is which is processed by the blast wave is more or less contained at the blast wave front itself therefore this was from mass balance let us try to look at energy balance whether I can draw any other information again let us be very clear let us do the energy balance only for the condition when the lead blast wave is strong that is when the Mach number of the shock front is greater than around 4 and let us see whether we can draw some physics out of it again we assume a power law profile we assume that well the density can be given by at the lead shock the density jumps the density decreases so also the value of pressure decreases so also the value of u decreases or let us let us try to put these things together in some form and make may be make an energy balance what is the energy balance now I can go straight to the case of the lead shock being at rs may be the value we are doing for an element a spherical shock the lead shock is at rs now we are considering a particular radius r over here we are considering an spherical segment of radius r with this dr let us assume that the lead shock wave travels with a velocity rs dot such that the Mach number is still high greater than around 4 such that we can still use those values which are fairly constant with which we got started now I want to calculate the energy over here the change in energy now initially I had this this was this particular segment before the lead shock process did this particular segment had an initial energy E0 it was quiescent that means there was no motion once the blast wave processes this particular small segment it has an it has now an energy let us say E which is over and above E0 has this value it also has some kinetic energy may be u square divided by 2 use the particle velocity once the blast wave progresses well we are presumed that there is some change the particle velocity here could be u u1 u2 u3 here it is u and the particle velocity is keeping changing we are looking at the particle velocity within the small element over here this is per unit mass energy per unit mass kinetic energy therefore per unit volume I say it is equal to rho E minus E0 let us presume that the value of the energy after the blast wave processes it is very much greater than E0 because we are talking in terms of a strong blast wave let us presume that E is greater than E0 these are assumptions we make therefore we tell ourselves the energy in the element is equal to per unit mass is equal to E plus u square by 2 well the internal energy is is equal to the molecular energy which is equal to Cv into T and this is per unit mass per unit volume is equal to rho into Cv into T plus u square divided by 2 and per unit volume this is the value the volume of the small spherical segment again is equal to 4 pi r square which is the surface area multiplied by dr over which the properties are constant and therefore the energy within this element d is equal to 4 pi r square into dr which is the volume into the value of rho into Cv T this is the internal energy of this particular volume this is per unit mass this should have also been rho plus rho into u square divided by 2 is the energy within this small volume. Just like we balance the mass in this volume and integrated out over here we will integrate out and find out that the total energy what is the total energy what is now available is the energy from the explosion because initially it was only E0 we are looking at the excess over here and therefore the energy d e if I were to integrate out between 0 to Rs should be the energy which is released by the explosion because what happens the energy released by the explosion is dispersed by the lead shock wave over this particular volume and therefore I get the value of E0 is equal to let us put it down over here 4 pi r square into dr this is the volume into rho into u square plus rho u square by 2 kinetic energy plus rho into Cv T over here and this is equal to the energy which is released which is E0. Now let us try to simplify this this becomes equal to 4 pi r square 4 pi into well r square depends on Rs I cannot take it out I get 4 pi into integral 0 to Rs. Now I get the value of rho u square well u depends on the shock front density I still keep it as rho u square divided by 2 plus I get Cv can I simplify Cv you know if I look at Cp minus Cv is equal to the specific gas constant Cv is equal to Cp by Cv is gamma minus 1 is equal to r or rather Cv is equal to r over gamma minus 1 therefore I put this instead of this I get rho into r divided by gamma minus 1 into T over here into r square in dr and now I now make I now note that I have Pv is equal to RT for a gas Pv is equal to mass into RT or P by rho is equal to RT if I have to substitute the value of P by rho instead of r and T product of r and T what is it I will get I will get P by rho over here and this rho and rho gets cancelled and what is it I get this is equal to 4 pi into 0 to Rs of rho u square divided by 2 plus I have P by rho rho gets cancelled P divided by gamma minus 1 into r square into dr which is equal to the energy release in the explosion this is the total value of energy release and what this energy release is it is dispersing through the medium and it is changing the value of density particle velocity and pressure in the particular medium. I want to be able to solve this and to be able to solve this I make a series of assumptions like I still consider my wave to be strong so that the conditions at the wave front are known in other words see we have already seen the following we told ourselves well the value if I were to consider the lead shock over here the density is let us say rho s it has jumped from the value of rho 0 to rho s and it keeps coming down over here therefore the density at any particular point rho by rho 0 can be written as equal to the value of rho by rho s into the value of rho by rho 0 which I can again write as some quantity let us say phi of r by Rs in other words see this is a particular value that means rho s by rho 0 such that I get rho by rho 0 rho by rho 0 can be written as some function of r by Rs similarly I can write p by rho 0 r s square at any particular point that is this is the shock front I am interested in a radius let us say r over here I can write it as equal to may be some function of r by Rs similarly the last term which I want to write I can also write u by Rs dot can also be written as some function of r by Rs mind you these could be expressed as power law but right now we say well it is a function of the initial condition it is a function of rho into Rs dot square it is a function of Rs dot over here and therefore what I will do in the next classes may be I will use these particular values substitute it into the energy equation and try to solve how the strong blast wave changes its characteristics or how the blast wave behaves when it propagates at high Mach number well then thank you that is about it.